We are asked to evaluate the indefinite integral of $\sin^2 x$. That is, we need to find $\int \sin^2 x \, dx$.

1. Problem Description

We are asked to evaluate the indefinite integral of $\sin^2 x$. That is, we need to find

$\int \sin^2 x \, dx$.

2. Solution Steps

We will use the trigonometric identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$ to simplify the integral.

$\int \sin^2 x \, dx = \int \frac{1 - \cos(2x)}{2} \, dx$

$= \frac{1}{2} \int (1 - \cos(2x)) \, dx$

$= \frac{1}{2} \left( \int 1 \, dx - \int \cos(2x) \, dx \right)$

$= \frac{1}{2} \left( x - \frac{1}{2} \sin(2x) \right) + C$

$= \frac{1}{2} x - \frac{1}{4} \sin(2x) + C$

We can verify this result by differentiating:

$\frac{d}{dx} \left( \frac{1}{2} x - \frac{1}{4} \sin(2x) + C \right) = \frac{1}{2} - \frac{1}{4} (2\cos(2x)) = \frac{1}{2} - \frac{1}{2} \cos(2x) = \frac{1 - \cos(2x)}{2} = \sin^2 x$.

3. Final Answer

$\int \sin^2 x \, dx = \frac{1}{2} x - \frac{1}{4} \sin(2x) + C$